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3 years ago

What is the specific heat of the solid phase? (Please see picture attached)

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

B.0.2 J/g°C

Explanation:

From the attached picture;

Heat attained in the solid phase is 200 Joules
Change in temperature is 50°C ( from 0°C to 50°C)
Mass of the solid is 20 g

We are required to determine the specific heat capacity of the substance;

We need to know that Quantity of heat is given by the product of mass,specific heat capacity and change in temperature.
That is; Q = mcΔT

Rearranging the formula;

c = Q ÷ mΔT

Therefore;

Specific heat = 200 J ÷ (20 g × 50°c)

= 0.2 J/g°C

Thus, the specific heat of the solid is 0.2 J/g°C

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. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.

Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

P1V1 / T1 = P2V2 / T2

-Solve for V2

V2 = P1V1T2 / T1P2

-Substitution

V2 = (6)(1)(252) / (295)(0.45)

- Simplification

V2 = 1512 / 132.75

- Result

V2 = 11.38 L

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3 years ago

Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas

Inessa05 [86]

The relative molecular mass of the gas : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

The relative molecular mass of the gas

Solution

Graham's Law

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas

Input the value :

$\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol$

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2 years ago

What are the transition metals

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3 years ago

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Cci(4) will be soluble in

Romashka-Z-Leto [24]

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3 years ago

Study this chemical reaction: (aq)(s)(s)(aq) Then, write balanced half-reactions describing the oxidation and reduction that hap

disa [49]

The question is incomplete, complete question is:

Study this chemical reaction:

$FeSO_4 (aq) + Zn (s)\rightarrow Fe (s) + ZnSO_4 (aq)$

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Oxidation:

Reduction:

Answer:

Oxidation taking place in given reaction :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction taking place in given reaction;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

$FeSO_4 (aq) + Zn (s) \rightarrow Fe (s) + ZnSO_4 (aq)$

In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.

Oxidation :

$Zn(s)\rightarrow Zn^{2+}+2e^-$

Reduction ;

$Fe^{2+}(aq)+2e^-\rightarrow Fe(s)$

6 0

3 years ago